Question

4) (16.28 of text) The prismatic beam in the figure has elastic modulus E, moment of inertia, I and length L. Determine the reaction forces at points A and B and determine the deflection as a function of x (16.29 of text). TW

I've solved the reaction forces, but I'm still trying to learn how to calculate the deflection.

Answer 1

Elastic modulus, E Moment of Inertia, I; length, 2 • origin is selected at A. statically indeterminete with degree of indeterminaty Stelpl The beam has sial, and has 3 reactions namely i moment at Á, vertical reaction at A * so the beam can be made statically determinate by taking anyone of reacts as redundant, I have taken moment at A (MA) as redundant at B . and vertical reaction W. AL mer w=won Statically indeterminete beam with SI=1 figl fig? ... making beam simply supported (determinate by taking momentat A as redundant I finding Reactions see fig 2 (1) Met moment at B PEMB=0 -MA TRAXL - (2x W x2) x 4 =0 . RA= MA + wote 1-0 cid Efyp=0 → Ratro = 4wol Ro=1 Wol - MA -Wall

step2 In this step we will find all reactions as well as determine the deflection as a function of a. * We will just derive the deflection curve equation Corigin as A) and use boundary conditions to get all the requirement of question . word origin A wort Bending Moment at a distance from A can be written as, Mr = RAK-MA (1 w x x Mr = (MA tuollu-MA -1 (W x) x. 2 & Mu= Man+Wohn-ma-world * Also we know that equation of elastic wave is EI dry Me > EI dy = MAX Two LX-MA-wors (i) Integrating with respect to die we will get Eile din = ((Ma x twok x- MA -Wome' Jude Eidst a Matt & Wols _ Max-went to - where C, is constant of integration 6L

cin Integrating u6 againt with respect to du Wolne _MAX-Woxta de 241. 12 Ely = MAX two 2 - Maxwell the 6L * So we have 2 equations ① & ② eq@ for slope (dup sloke) and eq@ for deflection (y-deflection) EI do = a 2 + who lost - Max-work on the Elya mo me twinx Max? _wo get tcxtc@ We also have 3 boundary conditions DALA, ALR=0, slope dy =o (fixed and see fig) using the 16) = 0+0=16=0 At A, deflection y-o >3=0,7=0 O using eg 6 o= otaxota [22=0 ② At B ie at UIL, deflection y=0 :- * After 6=0 & 62 = 0 egnation reduces to Atual, yzo

0= MAL two LX Mat - Wiktion * Using 0 &value of a RA= ( 7 wl) x twote = a wel * Using &value of Ma - Ro = Woh - ( 2 wolllat = 4 ush * Using eq6 & MA Ety = (wal) (L ) + Wire (24) -oors - 120L SEIY= 3 wolk - 7 / 5 us 2x² - wors 1201 Answer Equation of deflection as function of The equation for deflection is derived by considering 'y' axis upward so for downward deflection our equation will give neagative sign (just sign difference) if we want positive sign for downward deflection then we have to just make one change in above equation which is multiplying right hand side of equation by '-1'. For downward positive deflection, in derivation it will be (dzy/dx2 = -Mx) or multiply RHS of final eq. by '-1'. I Answer. moment at A, MA= 7 Woll 45 Reaction at A, RA= 9 Wola Reaction at B, RB = L Wolf